I´m getting all bangup with the maths!
If the cpu is as Q9550 (45m) and a P5q3 that as cpu gtl 0/2 and 1/3 with:
Vcore: 1.40
cpu gtl 0/2: auto
cpu gtl 1/3: auto
Vpll: 1.60
vfsb: 1.46
vdram: 2.10
NB gtl: auto
Vnb: 1.38
Vsb: 1.10
VPcie: 1.50
i cant get write the multi. Y= (vtt*0.670)-(vtt*0.630)
then Y+0.630= Z value that i will use put on the cpu gtl =0/2 AND cpu gtl= 1/3????
OR Y+0.630= cpu gtl 0/2 AND Y+0.670=cpu gtl 1/3 ??????
can someone PLEASE run the maths for me with steps because my values are all wrong....?!
Tks!
vox




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